The specific heat of water = 4200 J kg^{-1} K^{-1} and the latent heat of ice = 3.4 \times 10^{5}J kg^{-1}, 100 grams of ice at 0^{\circ}C is placed 200g of water at 25^{\circ}C. The amount of ice that will melt as the temperature of water reaches 0^{\circ}C is close to (in grams):
Option: 1 61.7
Option: 2 63.8
Option: 3 69.3
Option: 4 64.6

Answers (1)

Here the water will provide heat for ice to melt therefore
 

i.e \ \ \mathrm{m}_{\mathrm{w}} \mathrm{s}_{\mathrm{w}} \Delta \theta=\mathrm{m}_{\text {ice }} \mathrm{L}_{\text {ice }} \\ \Rightarrow \mathrm{m}_{\mathrm{ice}}=\frac{0.2 \times 4200 \times 25}{3.4 \times 10^{5}}$ $=0.0617 \mathrm{~kg}$ $=61.7 \mathrm{gm}

The remaining ice will remain un-melted.

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